Ultrafilter Extensions and the Standard Translation

Notes from Lecture 4: Ultraﬁlter Extensions and the Standard Translation

Eric Pacuit∗

March 1, 2012

1 Ultraﬁlter Extensions

Definition 1.1 (Ultrafilter) Let W be a non-empty set. An ultrafilter on W is a set u ⊆ ℘(W ) satisfying the following properties: 1. ∅ 6∈ u 2. If X,Y ∈ u then X ∩ Y ∈ u 3. If X ∈ u and X ⊆ Y then Y ∈ u. 4. For all X ⊆ W , either X ∈ u or X ∈ u (where X is the complement of X in W ) /

A collection u0 ⊆ ℘(W ) has the ﬁnite intersection property provided for each X,Y ∈ u0, X ∩ Y 6= ∅.

Theorem 1.2 (Ultrafilter Theorem) If a set u0 ⊆ ℘(W ) has the finite intersection property, then u0 can be extended to an ultrafilter over W (i.e., there is an ultrafilter u over W such that u0 ⊆ u.

Proof. Suppose that u0 has the ﬁnite intersection property. Then, consider the set u1 = {Z | there are ﬁnitely many sets X1,...Xk such that Z = X1 ∩ · · · ∩ Xk}.

That is, u1 is the set of finite intersections of sets from u0. Note that u0 ⊆ u1, since u0 has the finite intersection property, we have ∅ 6∈ u1, and by definition u1 is closed under finite intersections. Now, define u2 as follows: 0 u = {Y | there is a Z ∈ u1 such that Z ⊆ Y }

∗UMD, Philosophy. Webpage: ai.stanford.edu/∼epacuit, Email: [email protected]

1 0 0 We claim that u is a consistent filter: Y1,Y2 ∈ u then there is a Z1 ∈ u1 such that Z1 ⊆ Y1 and Z2 ∈ u1 such that Z2 ⊆ Y2. Then, since ∅ 6= Z1 ∩ Z2 ∈ u1, we have Z1 ∩ Z2 ⊆ Y1 ∩ Y2. Hence, Y1 ∩ Y2 ∈ u1. Also, if X ∈ u1 then there is a Z ∈ u1 such that Z ⊆ X. If X ⊆ Y , then Z ⊆ Y and so Y ∈ u1. Hence, u1 is a consistent filter. The next step is to show that u1 can be extended to an ultrafilter. This follows almost directly from Zorn’s Lemma1: Consider the set Z of all filters that extend u1. That is, Z = {v | u1 ⊆ v and v is a consistent filter}. Note that Z is partially-ordered under the ⊆-relation. Furthermore, the upper bound of any chain in Z (i.e., sequence of ultrafilters v0 ⊆ v1 ⊆ · · · ) is the union of all the filters in the chain. This collection of sets will be a consistent ultrafilter extending u1, and so is contained in Z. By Zorn’s Lemma, Z must contain a maximal element. This maximal element must be an ultrafilter (containing u1). qed

Let M = hW, R, V i be a Kripke model. Two functions are relevant to our analysis: • m : ℘(W ) → ℘(W ) deﬁned as m(X) = {w | there is a v such that wRv and v ∈ X}, and

• l : ℘(W ) → ℘(W ) deﬁned as l(X) = {w | for all v, if wRv then v ∈ X}.

Definition 1.3 (Ultrafilter Extension) An ultrafilter extension is a model ue(M) = hUf(W ),Rue,V uei where • Uf(W ) = {u | u is an ultrafilter over W },

• uRueu0 iﬀ for all X ⊆ W , if X ∈ u0 then m(X) ∈ u, and

• V ue(p) = {u | V (p) ∈ u}. /

Fact 1.4 In an ultraﬁlter extension ue(M) = hUf(W ),Rue,V uei, we have uRueu0 iﬀ {Y | l(Y ) ∈ u} ⊆ u0.

Proof. Left as an exercise. qed

Let M = hW, R, V i be a model . The truth map [[·]]M : L → ℘(W ) is deﬁned by induction on the structure of ϕ as follows:

• For p ∈ At, [[p]]M = V (p)

1Zorn’s Lemma states that in a partially ordered set P , if every chain has an upper bound in P , then P contains at least on maximal element. The proof of Zorn’s Lemma uses the Axiom of Choice (indeed, it is equivalent to the Axiom of Choice).

2 • [[¬ϕ]]M = W − [[ϕ]]M

• [[ϕ ∧ ψ]]M = [[ϕ]]M ∩ [[ψ]]M

• [[3ϕ]]M = m([[ϕ]]M) Lemma 1.5 For all models M = hW, R, V i, for all modal formulas ϕ, we have [[ϕ]]M ∈ u iﬀ ue(M), u |= ϕ. Proof. The proof is by induction on the structure of ϕ. Then we have,

Base case: ϕ is p ∈ At.

[[p]]M ∈ u iff V (p) ∈ u (Definition of [[·]]M) iff u ∈ V ue(p) (Definition of V ue) iff ue(M), u |= p (Definition of truth in a model)

2 Induction Hypothesis: For all ψ less complex than ϕ, [[ψ]]M ∈ u iﬀ ue(M), u |= ψ

Case 1: ϕ is ¬ψ:

[[¬ψ]]M ∈ u iff W − [[ψ]]M ∈ u (Definition of [[·]]M) iff [[ψ]]M 6∈ u (Properties of an ultrafilter) iff ue(M), u 6|= ψ (Induction Hypothesis) iff ue(M), u |= ¬ψ (Definition of truth)

Case 2: ϕ is ψ1 ∧ ψ2

[[ψ1 ∧ ψ2]]M ∈ u iff [[ψ1]]M ∩ [[ψ2]]M ∈ u (Definition of [[·]]M) iff [[ψ1]]M ∈ u and [[ψ2]]M ∈ u (Properties of ultrafilters) iff ue(M), u |= ψ1 and ue(M), u |= ψ2 (Induction hypothesis) iff ue(M), u |= ψ1 ∧ ψ2 (Definition of truth) Case 3: ϕ is 3ψ

Suppose that ue(M), u |= 3ψ. Then, there is a u0 ∈ Uf(W ) such that uRueu0 0 0 and ue(M), u |= ψ. By the induction hypothesis, [[ψ]]M ∈ u . By the deﬁnition ue of R , we have m([[ψ]]M) ∈ u, and so, [[3ψ]]M ∈ u. Thus, we have shown that ue(M), u |= 3ψ implies [[3ψ]]M ∈ u. Suppose that [[3ψ]]M ∈ u. We must show ue(M), u |= 3ψ. Consider the set

u0 = {Y | l(Y ) ∈ u} ∪ {[[ψ]]M} 2Less complex means that ψ contains fewer connectives.

3 We claim that u0 has the finite intersection property. We first show that {Y | l(Y ) ∈ u} is closed under finite intersections. It is enough to show that for any two sets Y1,Y2 ∈ {Y | l(Y ) ∈ u}, Y1 ∩ Y2 ∈ {Y | l(Y ) ∈ u} (why?). Suppose that Y1,Y2 ∈ {Y | l(Y ) ∈ u}. Note that l(Y1 ∩ Y2) = l(Y1) ∩ l(Y2). Then, since u is an ultrafilter and l(Y1), l(Y2) ∈ u, we have l(Y1 ∩ Y2) = l(Y1) ∩ l(Y2) ∈ u. Hence Y1 ∩ Y2 ∈ {Y | l(Y ) ∈ u}. Next we show that for any Z ∈ {Y | l(Y ) ∈ u}, we have Z ∩ [[ψ]]M 6= ∅. Choose an arbitrary Z such that l(Z) ∈ u. We will show Z ∩ [[ψ]]M 6= ∅. Since l(Z) ∈ u and [[3ψ]]M ∈ u, we have l(Z) ∩ [[3ψ]]M ∈ u, and so l(Z) ∩ [[3ψ]]M 6= ∅. Let w ∈ l(Z) ∩ [[3ψ]]M. Then, there is a v ∈ W such that M, v |= ψ. I.e., wRv and v ∈ [[ψ]]M. Since w ∈ l(Y ) and wRv, we have v ∈ Y . Hence, v ∈ Y ∩ [[ψ]]M. This implies that u0 has the finite intersection property (why?). 0 0 By the ultrafilter theorem, there is an ultrafilter u such that u0 ⊆ u . Since {Y | l(Y ) ∈ u} ⊆ u0, we have uRueu0. By the induction hypothesis, since 0 0 [[ψ]]M ∈ u , we have ue(M), u |= ψ. Hence, ue(M), u |= 3ψ. qed

Corollary 1.6 For all models M and states w in M, we have w ! uw, where uw is the principle ultraﬁlter generated by w.

Proof. Let M = hW, R, V i be a model and w ∈ W . The principle ultrafilter generated by w is uw = {X ⊆ W | w ∈ X}. Let ϕ be an arbitrary modal formula. We have M, w |= ϕ iff w ∈ [[ϕ]]M iff [[ϕ]]M ∈ uw iff ue(M), uw |= ϕ (the latter equivalence follows from the above Lemma). qed

Lemma 1.7 For all models M, ue(M) is m-saturated.

Proof. Suppose that ue(M) = hUf(W ),Rue,V uei an ultrafilter extension of some model M = hW, R, V i. Let u ∈ Uf(W ) be any state in ue(M) and Σ be a arbitrary set of modal formulas. Suppose that every finite subset of Σ is satisfiable at some successor of u (i.e., for each finite set ∆ ⊆ Σ, there is a ue V state v∆ ∈ W such that uR v∆ and ue(M), v∆ |= ∆). We must find a state v ∈ W such that uRuev and ue(M), v |= Σ (i.e., for each ψ ∈ Σ, ue(M), v |= ψ). Consider the set

v0 = {Y | l(Y ) ∈ u} ∪ {[[ψ]]M | ψ ∈ Σ}

We will show v0 has the finite intersection property. Since {Y | l(Y ) ∈ u} is T closed under finite intersections, it is enough to show that Y ∩ {[[ψ]]M | ψ ∈ T V ∆} 6= ∅ for some finite subset ∆ of Σ. Note that {[[ψ]]M | ψ ∈ ∆} = [[ ∆]]M. ue Recall that ∆ is satisfiable at some successor state v∆ of u. That is, uR v∆ V V and ue(M), v∆ |= ∆. By Lemma 1.5, this means [[ ∆]]M ∈ v∆. By the ue V V definition of R , we have m([[ ∆]]M) ∈ u. Hence, [[3 ∆]]M ∈ u. Since u is

4 V an ultrafilter, we have l(Y ) ∩ [[3 ∆]]M ∈ u. Hence, (since ∅ 6∈ u) there is a V V x ∈ l(Y ) ∩ [[3 ∆]]M. This implies there is a y such that xRy and y ∈ [[ ∆]]M. V Since xRy and x ∈ l(Y ), we have y ∈ Y . This means that y ∈ Y ∩ [[ ∆]]M. Hence, v0 has the finite intersection property. By the ultrafilter theorem there is an ultrafilter v such that v0 ⊆ v. By construction v is a successor u (i.e., uRuev) and by Lemma 1.5, we have for each ψ ∈ Σ, ue(M), v |= ψ. Hence, Σ is satisfiable is some successor state of u. qed

Theorem 1.8 (Bisimulation Somewhere Else Theorem) For all models M 0 0 0 0 and M , we have M, w ! M , w iﬀ ue(M), uw ↔ ue(M ), uw0 , where uw and 0 uw0 are the principle ultraﬁlters containing w and w respectively.

0 Proof. Suppose that ue(M), uw ↔ ue(M ), uw0 . Let ϕ be any model formula. We have

M, w |= ϕ iff ue(M), uw |= ϕ (Corollary 1.6) 0 iff ue(M ), uw0 |= ϕ (Bisimulation implies modal equivalence) iff M0, w0 |= ϕ (Corollary 1.6) 0 0 Suppose that M, w ! M , w . Then, by Corollary 1.6, we have ue(M), uw ! 0 0 0 0 M, w ! M , w ! ue(M ), uw0 . By Lemma 1.7, both ue(M) and ue(M ) are modally saturated. In modally saturated models, modal equivalence implies 0 bisimularity. Hence, ue(M), uw ↔ ue(M ), uw0 . qed

2 The Standard Translation

Let M = hW, R, V i be a Kripke model. The ﬁrst-order language L1 is built from a signature containing unary predicate symbols P x corresponding to each p ∈ At and a binary predicate symbol Rxy. The standard translation is deﬁned as follows:

Deﬁnition 2.1 (Standard Translation) The standard translation are functions stx : L → L1 deﬁned as follows:

stx(p) = P x stx(¬ϕ) = ¬stx(ϕ) stx(ϕ ∧ ψ) = stx(ϕ) ∧ stx(ψ) stx(2ϕ) = ∀y(Rxy → sty(ϕ)) stx(3ϕ) = ∃y(Rxy ∧ sty(ϕ)) ads /

5 Observation 2.2 Modal logic falls in the two-variable fragment of L1.

Proof. By carefully reusing bound variables, once can ensure that the translation of a modal formula uses only two variable. An example suﬃces to show how this works:

stx(33p) = ∃y(Rxy ∧ sty(3p)) = ∃y(Rxy ∧ (∃x(Ryx ∧ P x)))

qed

Lemma 2.3 Let M = hW, R, V i be a Kripke model. For each w ∈ W , M, w |= ϕ iﬀ M stx(ϕ)[x/w], where denotes truth of L1 in a model M (viewed as a ﬁrst-order structure).

Proof. The simple but instructive proof is left to the reader. qed

Theorem 2.4 (Van Benthem Characterization Theorem) A ﬁrst-order formula α(x) (in the appropriate language) is invariant for bisimulation iﬀ it is equivalent to the translation of a modal formula.